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A circle of constant radius \mathrm{'a'} passes through origin \mathrm{'O'} and cuts the axes of co-ordinates in points \mathrm{P} and \mathrm{Q}, then the equation of the locus of the foot of perpendicular from \mathrm{O} to \mathrm{PQ} is:

Option: 1

\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)\left(\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}\right)=4 \mathrm{a}^{2}


Option: 2

\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}\left(\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}\right)=\mathrm{a}^{2}


Option: 3

\mathrm{\left(x^{2}+y^{2}\right)^{2}\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right)=4 a^{2}}


Option: 4

\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)\left(\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}\right)=\mathrm{a}^{2}


Answers (1)

best_answer

Equation of line PQ is

\mathrm{y}-\mathrm{k}=-\frac{\mathrm{h}}{\mathrm{k}}(\mathrm{x}-\mathrm{h}) \quad \Rightarrow \quad \mathrm{hx}+\mathrm{ky}=\mathrm{h}^{2}+\mathrm{k}^{2} ; \text { also } 2 \mathrm{a}=\sqrt{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}} \Rightarrow \mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}=4 \mathrm{a}^{2}

\therefore \quad \text { points } \mathrm{Q}\left(\frac{\mathrm{h}^{2}+\mathrm{k}^{2}}{\mathrm{~h}}, 0\right) \text { and } \mathrm{P}\left(0, \frac{\mathrm{h}^{2}+\mathrm{k}^{2}}{\mathrm{k}}\right)
\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}\left(\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}\right)=4 \mathrm{a}^{2}

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Nehul

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